Random number generation in $3D$

Question

I have problem regarding random number generation. Suppose I have disc of radius $r$. $$ \begin{align} x&=r\cos(\theta)\\ y&=r\sin(\theta)\\ z&=0 \end{align} $$ I rotate the co-ordinate. So new co-ordinate is $$ \begin{align} X&=x\cos(\phi)+z\sin(\phi)\\ Y&=y\\ Z&=-x\sin(\phi)+ z\cos(\phi) \end{align} $$ I want to generate $r$, $\theta$ and $\phi$ by random number such that I get equal probability in all space. I used $$ \begin{align} r&=r_{\text{max}}\sqrt{\mathrm{random}(N)}\\ \theta&=2\pi\,\mathrm{random}(N)\\ \phi&=\phi_{\text{min}}+\mathrm{random}(N)*(\phi_{\text{max}}-\phi_{\text{min}}) \end{align} $$ But it does not work I think. Can anybody help me?

Answer 1

If you want your coordinates uniform over a sphere with radius $R$, you want the density of $r$ to be proportional to $r^2$. Section 7.2 of Numerical Recipes describes how to transform uniform deviates on $[0,1]$ to a desired distribution when you can integrate the inverse function. here you want $p(y)dy=r^2\ dy$, so need to find a function $y(x)$ such that $\left | \frac {dx}{dy} \right |=y^2$, which we can see to be $x=\frac 13y^3, y=\sqrt[3]{3x}$ so your $r$ coordinate is $R\sqrt[3]{x}$ for $x$ a standard random. $\phi$ is easy-you want it uniform on $[0,2\pi]$ so just multiply a random by $2\pi$. For $\theta$, you want the density proportional to $\sin \theta$, so you need $\left | \frac {dx}{dy} \right |=\sin y$, then $x=\cos y$ and your relation is $y=\arccos (2x-1)$

Answer 2

Suppose a random variable $X$ has the Cumulative Distribution Function $\Phi(t)=P(X\le t)$:

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The probability of $X$ being within the base of the green region, is the height of the red region. Thus, $\Phi(X)$ is uniformly distributed along the vertical axis, $[0,1]$. Therefore, $\Phi^{-1}(U)$ has the same distribution as $X$ where $U$ is uniformly distributed in $[0,1]$.

As long as the region is radial, the CDF of being within $t$ of the origin is $$ \Phi(t)=\frac{t^3}{r^3} $$ Thus, we get the proper distribution of the distance from the origin with $$ r\,u^{1/3}\tag{1} $$ where $u$ is uniform in $[0,1]$.

Below, we look at shapes that are radially symmetric, and we need to use the signed radius $$ r\,(2u-1)^{1/3}\tag{2} $$

As shown in this answer, the the Lambert cylindrical projection preserves area. Thus. to generate a uniformly distributed point on the surface of the sphere, we can choose $z$ to be uniformly distributed in whatever range we wish, and choose $x$ and $y$ uniformly around the (possibly partial) circle.

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If the region is supposed to look like this between polar angles $\phi_{\text{min}}$ and $\phi_{\text{max}}$:

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Then we would take $$ \begin{align} z&=\cos(\phi_{\text{max}})+(\cos(\phi_{\text{min}})-\cos(\phi_{\text{max}}))v\\ y&=\sin(2\pi w)\sqrt{1-z^2}\\ x&=\cos(2\pi w)\sqrt{1-z^2} \end{align}\tag{3} $$ where $v$ and $w$ are uniform in $[0,1]$.

To get the points in the sphere, multiply the points on the sphere from $(3)$ by the radius from $(2)$.

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If the region is supposed to look like this between equatorial angles $\phi_{\text{min}}$ and $\phi_{\text{max}}$:

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Then we would take $$ \begin{align} z&=2v-1\\ y&=\sin(\phi_{\text{min}}+(\phi_{\text{max}}-\phi_{\text{min}})w)\sqrt{1-z^2}\\ x&=\cos(\phi_{\text{min}}+(\phi_{\text{max}}-\phi_{\text{min}})w)\sqrt{1-z^2} \end{align}\tag{4} $$ where $v$ and $w$ are uniform in $[0,1]$.

To get the points in the sphere, multiply the points on the sphere from $(4)$ by the radius from $(2)$.

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As verified by the author, the situation is supposed to be as generated by $(4)$. Using the coordinates given in the question, we get $$ \begin{align} r&=r_{\text{max}}(2u-1)^{1/3}\\ \theta&=\sin^{-1}(2v-1)\\ \phi&=\phi_{\text{min}}+(\phi_{\text{max}}-\phi_{\text{min}})w \end{align}\tag{5} $$ where $u,v,w$ are uniform in $[0,1]$. If you don't like negative $r$, when $r\lt0$, use $$ \begin{align} r'&=-r\\ \theta'&=\theta+\pi\\ \phi'&=\phi \end{align}\tag{6} $$

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10000 Points: Generated using $(2)$ and $(3)$ (top is approaching; bottom is receding)

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10000 Points: Generated using $(2)$ and $(4)$ (top is approaching; bottom is receding)

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